有关最长公共子序列算法只是一个简单的问题。 我这样做，你需要生成序列的部分如下：

Just a quick question about the longest Common subsequence algorithm. I have done the part where you need to generate the subsequence as follow:

public int[][] lcsLength(char[] input1, char[] input2) {
    int[][] opt = new int[M][N];
    for (int i = 1; i < input1.length; i++) {
        for (int j = 1; j < input2.length; j++) {
            if (input1[i] == input2[j]) {
                opt[i][j] = opt[i - 1][j - 1] + 1;
            } else {
                opt[i][j] = Math.max(opt[i][j - 1], opt[i - 1][j]);
            }
        }
    }
    return opt;
}

和printDiff功能如下：

and the printDiff function as follow:

  private static void printDiff(int[][] opt,String x,String y,int i, int j) {

    if(i>0 &&j>0 && x.charAt(i-1)==y.charAt(j-1)){
    printDiff(i-1,j-1);
    System.out.println(x.charAt(i-1));
    }
    else{
        if(j>0&&(i==0||opt[i][j-1]>=opt[i-1][j])){
        printDiff(i-1,j-1);
         System.out.println("-"+y.charAt(j-1));
        }
        else if(i>0&&(j==0|| opt[i][j-1]<=opt[i-1][j])){
        printDiff(i-1,j-1);
         System.out.println(x.charAt(i-1));
        }
    }

}

然后，如果我用这个作为参数：

And then if I use this as parameters:

String input1="ABCDE"
String input2="ACDC"
int i=input1.length()
int j=input2.length()

产生选择矩阵lcsLength（）后，我想，竟被printdiff给我：

after generating the opt matrix with lcsLength() I wish that printdiff woul give me :

ABCDE-
A-CD-C

而是我得到的：

but instead I get:

ABCDE-
ABCD-C

这是我做错了什么会帮助我很多的任何想法

any ideas on what I did wrong would help me a lot

谢谢 洛朗

推荐答案

不知道，如果它是一个相关的问题，但我觉得你的LCS code应该是：

Don't know if it's a related issue, but I think your LCS code should be:

public int[][] lcsLength(char[] input1, char[] input2) {
    int[][] opt = new int[input1.length+1][input2.length+1];
    for (int i = 1; i <= input1.length; i++) {
        for (int j = 1; j <= input2.length; j++) {
            if (input1[i-1] == input2[j-1]) {
                opt[i][j] = opt[i - 1][j - 1] + 1;
            } else {
                opt[i][j] = Math.max(opt[i][j - 1], opt[i - 1][j]);
            }
        }
    }
    return opt;
}